3.240 \(\int \frac {c+d x^3+e x^6+f x^9}{x^2 (a+b x^3)} \, dx\)

Optimal. Leaf size=227 \[ -\frac {\log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right ) \left (a^3 (-f)+a^2 b e-a b^2 d+b^3 c\right )}{6 a^{4/3} b^{8/3}}+\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right ) \left (a^3 (-f)+a^2 b e-a b^2 d+b^3 c\right )}{3 a^{4/3} b^{8/3}}+\frac {\tan ^{-1}\left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right ) \left (a^3 (-f)+a^2 b e-a b^2 d+b^3 c\right )}{\sqrt {3} a^{4/3} b^{8/3}}+\frac {x^2 (b e-a f)}{2 b^2}-\frac {c}{a x}+\frac {f x^5}{5 b} \]

[Out]

-c/a/x+1/2*(-a*f+b*e)*x^2/b^2+1/5*f*x^5/b+1/3*(-a^3*f+a^2*b*e-a*b^2*d+b^3*c)*ln(a^(1/3)+b^(1/3)*x)/a^(4/3)/b^(
8/3)-1/6*(-a^3*f+a^2*b*e-a*b^2*d+b^3*c)*ln(a^(2/3)-a^(1/3)*b^(1/3)*x+b^(2/3)*x^2)/a^(4/3)/b^(8/3)+1/3*(-a^3*f+
a^2*b*e-a*b^2*d+b^3*c)*arctan(1/3*(a^(1/3)-2*b^(1/3)*x)/a^(1/3)*3^(1/2))/a^(4/3)/b^(8/3)*3^(1/2)

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Rubi [A]  time = 0.19, antiderivative size = 227, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.233, Rules used = {1834, 292, 31, 634, 617, 204, 628} \[ -\frac {\log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right ) \left (a^2 b e+a^3 (-f)-a b^2 d+b^3 c\right )}{6 a^{4/3} b^{8/3}}+\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right ) \left (a^2 b e+a^3 (-f)-a b^2 d+b^3 c\right )}{3 a^{4/3} b^{8/3}}+\frac {\tan ^{-1}\left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right ) \left (a^2 b e+a^3 (-f)-a b^2 d+b^3 c\right )}{\sqrt {3} a^{4/3} b^{8/3}}+\frac {x^2 (b e-a f)}{2 b^2}-\frac {c}{a x}+\frac {f x^5}{5 b} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x^3 + e*x^6 + f*x^9)/(x^2*(a + b*x^3)),x]

[Out]

-(c/(a*x)) + ((b*e - a*f)*x^2)/(2*b^2) + (f*x^5)/(5*b) + ((b^3*c - a*b^2*d + a^2*b*e - a^3*f)*ArcTan[(a^(1/3)
- 2*b^(1/3)*x)/(Sqrt[3]*a^(1/3))])/(Sqrt[3]*a^(4/3)*b^(8/3)) + ((b^3*c - a*b^2*d + a^2*b*e - a^3*f)*Log[a^(1/3
) + b^(1/3)*x])/(3*a^(4/3)*b^(8/3)) - ((b^3*c - a*b^2*d + a^2*b*e - a^3*f)*Log[a^(2/3) - a^(1/3)*b^(1/3)*x + b
^(2/3)*x^2])/(6*a^(4/3)*b^(8/3))

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 292

Int[(x_)/((a_) + (b_.)*(x_)^3), x_Symbol] :> -Dist[(3*Rt[a, 3]*Rt[b, 3])^(-1), Int[1/(Rt[a, 3] + Rt[b, 3]*x),
x], x] + Dist[1/(3*Rt[a, 3]*Rt[b, 3]), Int[(Rt[a, 3] + Rt[b, 3]*x)/(Rt[a, 3]^2 - Rt[a, 3]*Rt[b, 3]*x + Rt[b, 3
]^2*x^2), x], x] /; FreeQ[{a, b}, x]

Rule 617

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[(a*c)/b^2]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + (2*c*x)/b], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 634

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[(2*c*d - b*e)/(2*c), Int[1/(a +
b*x + c*x^2), x], x] + Dist[e/(2*c), Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] &
& NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] &&  !NiceSqrtQ[b^2 - 4*a*c]

Rule 1834

Int[((Pq_)*((c_.)*(x_))^(m_.))/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Int[ExpandIntegrand[((c*x)^m*Pq)/(a + b*
x^n), x], x] /; FreeQ[{a, b, c, m}, x] && PolyQ[Pq, x] && IntegerQ[n] &&  !IGtQ[m, 0]

Rubi steps

\begin {align*} \int \frac {c+d x^3+e x^6+f x^9}{x^2 \left (a+b x^3\right )} \, dx &=\int \left (\frac {c}{a x^2}+\frac {(b e-a f) x}{b^2}+\frac {f x^4}{b}+\frac {\left (-b^3 c+a b^2 d-a^2 b e+a^3 f\right ) x}{a b^2 \left (a+b x^3\right )}\right ) \, dx\\ &=-\frac {c}{a x}+\frac {(b e-a f) x^2}{2 b^2}+\frac {f x^5}{5 b}+\frac {\left (-b^3 c+a b^2 d-a^2 b e+a^3 f\right ) \int \frac {x}{a+b x^3} \, dx}{a b^2}\\ &=-\frac {c}{a x}+\frac {(b e-a f) x^2}{2 b^2}+\frac {f x^5}{5 b}+\frac {\left (b^3 c-a b^2 d+a^2 b e-a^3 f\right ) \int \frac {1}{\sqrt [3]{a}+\sqrt [3]{b} x} \, dx}{3 a^{4/3} b^{7/3}}-\frac {\left (b^3 c-a b^2 d+a^2 b e-a^3 f\right ) \int \frac {\sqrt [3]{a}+\sqrt [3]{b} x}{a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2} \, dx}{3 a^{4/3} b^{7/3}}\\ &=-\frac {c}{a x}+\frac {(b e-a f) x^2}{2 b^2}+\frac {f x^5}{5 b}+\frac {\left (b^3 c-a b^2 d+a^2 b e-a^3 f\right ) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 a^{4/3} b^{8/3}}-\frac {\left (b^3 c-a b^2 d+a^2 b e-a^3 f\right ) \int \frac {-\sqrt [3]{a} \sqrt [3]{b}+2 b^{2/3} x}{a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2} \, dx}{6 a^{4/3} b^{8/3}}-\frac {\left (b^3 c-a b^2 d+a^2 b e-a^3 f\right ) \int \frac {1}{a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2} \, dx}{2 a b^{7/3}}\\ &=-\frac {c}{a x}+\frac {(b e-a f) x^2}{2 b^2}+\frac {f x^5}{5 b}+\frac {\left (b^3 c-a b^2 d+a^2 b e-a^3 f\right ) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 a^{4/3} b^{8/3}}-\frac {\left (b^3 c-a b^2 d+a^2 b e-a^3 f\right ) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{6 a^{4/3} b^{8/3}}-\frac {\left (b^3 c-a b^2 d+a^2 b e-a^3 f\right ) \operatorname {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1-\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a}}\right )}{a^{4/3} b^{8/3}}\\ &=-\frac {c}{a x}+\frac {(b e-a f) x^2}{2 b^2}+\frac {f x^5}{5 b}+\frac {\left (b^3 c-a b^2 d+a^2 b e-a^3 f\right ) \tan ^{-1}\left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right )}{\sqrt {3} a^{4/3} b^{8/3}}+\frac {\left (b^3 c-a b^2 d+a^2 b e-a^3 f\right ) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 a^{4/3} b^{8/3}}-\frac {\left (b^3 c-a b^2 d+a^2 b e-a^3 f\right ) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{6 a^{4/3} b^{8/3}}\\ \end {align*}

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Mathematica [A]  time = 0.21, size = 224, normalized size = 0.99 \[ \frac {15 a^{4/3} b^{2/3} x^3 (b e-a f)+6 a^{4/3} b^{5/3} f x^6+10 x \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right ) \left (a^3 (-f)+a^2 b e-a b^2 d+b^3 c\right )+10 \sqrt {3} x \tan ^{-1}\left (\frac {1-\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a}}}{\sqrt {3}}\right ) \left (a^3 (-f)+a^2 b e-a b^2 d+b^3 c\right )-5 x \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right ) \left (a^3 (-f)+a^2 b e-a b^2 d+b^3 c\right )-30 \sqrt [3]{a} b^{8/3} c}{30 a^{4/3} b^{8/3} x} \]

Antiderivative was successfully verified.

[In]

Integrate[(c + d*x^3 + e*x^6 + f*x^9)/(x^2*(a + b*x^3)),x]

[Out]

(-30*a^(1/3)*b^(8/3)*c + 15*a^(4/3)*b^(2/3)*(b*e - a*f)*x^3 + 6*a^(4/3)*b^(5/3)*f*x^6 + 10*Sqrt[3]*(b^3*c - a*
b^2*d + a^2*b*e - a^3*f)*x*ArcTan[(1 - (2*b^(1/3)*x)/a^(1/3))/Sqrt[3]] + 10*(b^3*c - a*b^2*d + a^2*b*e - a^3*f
)*x*Log[a^(1/3) + b^(1/3)*x] - 5*(b^3*c - a*b^2*d + a^2*b*e - a^3*f)*x*Log[a^(2/3) - a^(1/3)*b^(1/3)*x + b^(2/
3)*x^2])/(30*a^(4/3)*b^(8/3)*x)

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fricas [A]  time = 0.64, size = 560, normalized size = 2.47 \[ \left [\frac {6 \, a^{2} b^{3} f x^{6} - 30 \, a b^{4} c + 15 \, {\left (a^{2} b^{3} e - a^{3} b^{2} f\right )} x^{3} - 15 \, \sqrt {\frac {1}{3}} {\left (a b^{4} c - a^{2} b^{3} d + a^{3} b^{2} e - a^{4} b f\right )} x \sqrt {\frac {\left (-a b^{2}\right )^{\frac {1}{3}}}{a}} \log \left (\frac {2 \, b^{2} x^{3} - a b + 3 \, \sqrt {\frac {1}{3}} {\left (a b x + 2 \, \left (-a b^{2}\right )^{\frac {2}{3}} x^{2} + \left (-a b^{2}\right )^{\frac {1}{3}} a\right )} \sqrt {\frac {\left (-a b^{2}\right )^{\frac {1}{3}}}{a}} - 3 \, \left (-a b^{2}\right )^{\frac {2}{3}} x}{b x^{3} + a}\right ) - 5 \, {\left (b^{3} c - a b^{2} d + a^{2} b e - a^{3} f\right )} \left (-a b^{2}\right )^{\frac {2}{3}} x \log \left (b^{2} x^{2} + \left (-a b^{2}\right )^{\frac {1}{3}} b x + \left (-a b^{2}\right )^{\frac {2}{3}}\right ) + 10 \, {\left (b^{3} c - a b^{2} d + a^{2} b e - a^{3} f\right )} \left (-a b^{2}\right )^{\frac {2}{3}} x \log \left (b x - \left (-a b^{2}\right )^{\frac {1}{3}}\right )}{30 \, a^{2} b^{4} x}, \frac {6 \, a^{2} b^{3} f x^{6} - 30 \, a b^{4} c + 15 \, {\left (a^{2} b^{3} e - a^{3} b^{2} f\right )} x^{3} - 30 \, \sqrt {\frac {1}{3}} {\left (a b^{4} c - a^{2} b^{3} d + a^{3} b^{2} e - a^{4} b f\right )} x \sqrt {-\frac {\left (-a b^{2}\right )^{\frac {1}{3}}}{a}} \arctan \left (\frac {\sqrt {\frac {1}{3}} {\left (2 \, b x + \left (-a b^{2}\right )^{\frac {1}{3}}\right )} \sqrt {-\frac {\left (-a b^{2}\right )^{\frac {1}{3}}}{a}}}{b}\right ) - 5 \, {\left (b^{3} c - a b^{2} d + a^{2} b e - a^{3} f\right )} \left (-a b^{2}\right )^{\frac {2}{3}} x \log \left (b^{2} x^{2} + \left (-a b^{2}\right )^{\frac {1}{3}} b x + \left (-a b^{2}\right )^{\frac {2}{3}}\right ) + 10 \, {\left (b^{3} c - a b^{2} d + a^{2} b e - a^{3} f\right )} \left (-a b^{2}\right )^{\frac {2}{3}} x \log \left (b x - \left (-a b^{2}\right )^{\frac {1}{3}}\right )}{30 \, a^{2} b^{4} x}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x^9+e*x^6+d*x^3+c)/x^2/(b*x^3+a),x, algorithm="fricas")

[Out]

[1/30*(6*a^2*b^3*f*x^6 - 30*a*b^4*c + 15*(a^2*b^3*e - a^3*b^2*f)*x^3 - 15*sqrt(1/3)*(a*b^4*c - a^2*b^3*d + a^3
*b^2*e - a^4*b*f)*x*sqrt((-a*b^2)^(1/3)/a)*log((2*b^2*x^3 - a*b + 3*sqrt(1/3)*(a*b*x + 2*(-a*b^2)^(2/3)*x^2 +
(-a*b^2)^(1/3)*a)*sqrt((-a*b^2)^(1/3)/a) - 3*(-a*b^2)^(2/3)*x)/(b*x^3 + a)) - 5*(b^3*c - a*b^2*d + a^2*b*e - a
^3*f)*(-a*b^2)^(2/3)*x*log(b^2*x^2 + (-a*b^2)^(1/3)*b*x + (-a*b^2)^(2/3)) + 10*(b^3*c - a*b^2*d + a^2*b*e - a^
3*f)*(-a*b^2)^(2/3)*x*log(b*x - (-a*b^2)^(1/3)))/(a^2*b^4*x), 1/30*(6*a^2*b^3*f*x^6 - 30*a*b^4*c + 15*(a^2*b^3
*e - a^3*b^2*f)*x^3 - 30*sqrt(1/3)*(a*b^4*c - a^2*b^3*d + a^3*b^2*e - a^4*b*f)*x*sqrt(-(-a*b^2)^(1/3)/a)*arcta
n(sqrt(1/3)*(2*b*x + (-a*b^2)^(1/3))*sqrt(-(-a*b^2)^(1/3)/a)/b) - 5*(b^3*c - a*b^2*d + a^2*b*e - a^3*f)*(-a*b^
2)^(2/3)*x*log(b^2*x^2 + (-a*b^2)^(1/3)*b*x + (-a*b^2)^(2/3)) + 10*(b^3*c - a*b^2*d + a^2*b*e - a^3*f)*(-a*b^2
)^(2/3)*x*log(b*x - (-a*b^2)^(1/3)))/(a^2*b^4*x)]

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giac [A]  time = 0.18, size = 269, normalized size = 1.19 \[ -\frac {\sqrt {3} {\left (b^{3} c - a b^{2} d - a^{3} f + a^{2} b e\right )} \arctan \left (\frac {\sqrt {3} {\left (2 \, x + \left (-\frac {a}{b}\right )^{\frac {1}{3}}\right )}}{3 \, \left (-\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{3 \, \left (-a b^{2}\right )^{\frac {1}{3}} a b^{2}} - \frac {c}{a x} + \frac {{\left (b^{3} c - a b^{2} d - a^{3} f + a^{2} b e\right )} \log \left (x^{2} + x \left (-\frac {a}{b}\right )^{\frac {1}{3}} + \left (-\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 \, \left (-a b^{2}\right )^{\frac {1}{3}} a b^{2}} + \frac {{\left (b^{3} c \left (-\frac {a}{b}\right )^{\frac {1}{3}} - a b^{2} d \left (-\frac {a}{b}\right )^{\frac {1}{3}} - a^{3} f \left (-\frac {a}{b}\right )^{\frac {1}{3}} + a^{2} b \left (-\frac {a}{b}\right )^{\frac {1}{3}} e\right )} \left (-\frac {a}{b}\right )^{\frac {1}{3}} \log \left ({\left | x - \left (-\frac {a}{b}\right )^{\frac {1}{3}} \right |}\right )}{3 \, a^{2} b^{2}} + \frac {2 \, b^{4} f x^{5} - 5 \, a b^{3} f x^{2} + 5 \, b^{4} x^{2} e}{10 \, b^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x^9+e*x^6+d*x^3+c)/x^2/(b*x^3+a),x, algorithm="giac")

[Out]

-1/3*sqrt(3)*(b^3*c - a*b^2*d - a^3*f + a^2*b*e)*arctan(1/3*sqrt(3)*(2*x + (-a/b)^(1/3))/(-a/b)^(1/3))/((-a*b^
2)^(1/3)*a*b^2) - c/(a*x) + 1/6*(b^3*c - a*b^2*d - a^3*f + a^2*b*e)*log(x^2 + x*(-a/b)^(1/3) + (-a/b)^(2/3))/(
(-a*b^2)^(1/3)*a*b^2) + 1/3*(b^3*c*(-a/b)^(1/3) - a*b^2*d*(-a/b)^(1/3) - a^3*f*(-a/b)^(1/3) + a^2*b*(-a/b)^(1/
3)*e)*(-a/b)^(1/3)*log(abs(x - (-a/b)^(1/3)))/(a^2*b^2) + 1/10*(2*b^4*f*x^5 - 5*a*b^3*f*x^2 + 5*b^4*x^2*e)/b^5

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maple [B]  time = 0.05, size = 419, normalized size = 1.85 \[ \frac {f \,x^{5}}{5 b}-\frac {a f \,x^{2}}{2 b^{2}}+\frac {e \,x^{2}}{2 b}+\frac {\sqrt {3}\, a^{2} f \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x}{\left (\frac {a}{b}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{3 \left (\frac {a}{b}\right )^{\frac {1}{3}} b^{3}}-\frac {a^{2} f \ln \left (x +\left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{3 \left (\frac {a}{b}\right )^{\frac {1}{3}} b^{3}}+\frac {a^{2} f \ln \left (x^{2}-\left (\frac {a}{b}\right )^{\frac {1}{3}} x +\left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 \left (\frac {a}{b}\right )^{\frac {1}{3}} b^{3}}-\frac {\sqrt {3}\, a e \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x}{\left (\frac {a}{b}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{3 \left (\frac {a}{b}\right )^{\frac {1}{3}} b^{2}}+\frac {a e \ln \left (x +\left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{3 \left (\frac {a}{b}\right )^{\frac {1}{3}} b^{2}}-\frac {a e \ln \left (x^{2}-\left (\frac {a}{b}\right )^{\frac {1}{3}} x +\left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 \left (\frac {a}{b}\right )^{\frac {1}{3}} b^{2}}-\frac {\sqrt {3}\, c \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x}{\left (\frac {a}{b}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{3 \left (\frac {a}{b}\right )^{\frac {1}{3}} a}+\frac {c \ln \left (x +\left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{3 \left (\frac {a}{b}\right )^{\frac {1}{3}} a}-\frac {c \ln \left (x^{2}-\left (\frac {a}{b}\right )^{\frac {1}{3}} x +\left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 \left (\frac {a}{b}\right )^{\frac {1}{3}} a}+\frac {\sqrt {3}\, d \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x}{\left (\frac {a}{b}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{3 \left (\frac {a}{b}\right )^{\frac {1}{3}} b}-\frac {d \ln \left (x +\left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{3 \left (\frac {a}{b}\right )^{\frac {1}{3}} b}+\frac {d \ln \left (x^{2}-\left (\frac {a}{b}\right )^{\frac {1}{3}} x +\left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 \left (\frac {a}{b}\right )^{\frac {1}{3}} b}-\frac {c}{a x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((f*x^9+e*x^6+d*x^3+c)/x^2/(b*x^3+a),x)

[Out]

1/5/b*f*x^5-1/2/b^2*x^2*a*f+1/2*e*x^2/b-1/3*a^2/b^3/(a/b)^(1/3)*ln(x+(a/b)^(1/3))*f+1/3*a/b^2/(a/b)^(1/3)*ln(x
+(a/b)^(1/3))*e-1/3/b/(a/b)^(1/3)*ln(x+(a/b)^(1/3))*d+1/3/a/(a/b)^(1/3)*ln(x+(a/b)^(1/3))*c+1/6*a^2/b^3/(a/b)^
(1/3)*ln(x^2-(a/b)^(1/3)*x+(a/b)^(2/3))*f-1/6*a/b^2/(a/b)^(1/3)*ln(x^2-(a/b)^(1/3)*x+(a/b)^(2/3))*e+1/6/b/(a/b
)^(1/3)*ln(x^2-(a/b)^(1/3)*x+(a/b)^(2/3))*d-1/6/a/(a/b)^(1/3)*ln(x^2-(a/b)^(1/3)*x+(a/b)^(2/3))*c+1/3*a^2/b^3*
3^(1/2)/(a/b)^(1/3)*arctan(1/3*3^(1/2)*(2/(a/b)^(1/3)*x-1))*f-1/3*a/b^2*3^(1/2)/(a/b)^(1/3)*arctan(1/3*3^(1/2)
*(2/(a/b)^(1/3)*x-1))*e+1/3/b*3^(1/2)/(a/b)^(1/3)*arctan(1/3*3^(1/2)*(2/(a/b)^(1/3)*x-1))*d-1/3/a*3^(1/2)/(a/b
)^(1/3)*arctan(1/3*3^(1/2)*(2/(a/b)^(1/3)*x-1))*c-1/a*c/x

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maxima [A]  time = 2.96, size = 217, normalized size = 0.96 \[ \frac {2 \, b f x^{5} + 5 \, {\left (b e - a f\right )} x^{2}}{10 \, b^{2}} - \frac {c}{a x} - \frac {\sqrt {3} {\left (b^{3} c - a b^{2} d + a^{2} b e - a^{3} f\right )} \arctan \left (\frac {\sqrt {3} {\left (2 \, x - \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}}{3 \, \left (\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{3 \, a b^{3} \left (\frac {a}{b}\right )^{\frac {1}{3}}} - \frac {{\left (b^{3} c - a b^{2} d + a^{2} b e - a^{3} f\right )} \log \left (x^{2} - x \left (\frac {a}{b}\right )^{\frac {1}{3}} + \left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 \, a b^{3} \left (\frac {a}{b}\right )^{\frac {1}{3}}} + \frac {{\left (b^{3} c - a b^{2} d + a^{2} b e - a^{3} f\right )} \log \left (x + \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{3 \, a b^{3} \left (\frac {a}{b}\right )^{\frac {1}{3}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x^9+e*x^6+d*x^3+c)/x^2/(b*x^3+a),x, algorithm="maxima")

[Out]

1/10*(2*b*f*x^5 + 5*(b*e - a*f)*x^2)/b^2 - c/(a*x) - 1/3*sqrt(3)*(b^3*c - a*b^2*d + a^2*b*e - a^3*f)*arctan(1/
3*sqrt(3)*(2*x - (a/b)^(1/3))/(a/b)^(1/3))/(a*b^3*(a/b)^(1/3)) - 1/6*(b^3*c - a*b^2*d + a^2*b*e - a^3*f)*log(x
^2 - x*(a/b)^(1/3) + (a/b)^(2/3))/(a*b^3*(a/b)^(1/3)) + 1/3*(b^3*c - a*b^2*d + a^2*b*e - a^3*f)*log(x + (a/b)^
(1/3))/(a*b^3*(a/b)^(1/3))

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mupad [B]  time = 5.37, size = 204, normalized size = 0.90 \[ x^2\,\left (\frac {e}{2\,b}-\frac {a\,f}{2\,b^2}\right )-\frac {c}{a\,x}+\frac {f\,x^5}{5\,b}+\frac {\ln \left (b^{1/3}\,x+a^{1/3}\right )\,\left (-f\,a^3+e\,a^2\,b-d\,a\,b^2+c\,b^3\right )}{3\,a^{4/3}\,b^{8/3}}-\frac {\ln \left (2\,b^{1/3}\,x-a^{1/3}+\sqrt {3}\,a^{1/3}\,1{}\mathrm {i}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (-f\,a^3+e\,a^2\,b-d\,a\,b^2+c\,b^3\right )}{3\,a^{4/3}\,b^{8/3}}+\frac {\ln \left (a^{1/3}-2\,b^{1/3}\,x+\sqrt {3}\,a^{1/3}\,1{}\mathrm {i}\right )\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (-f\,a^3+e\,a^2\,b-d\,a\,b^2+c\,b^3\right )}{3\,a^{4/3}\,b^{8/3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c + d*x^3 + e*x^6 + f*x^9)/(x^2*(a + b*x^3)),x)

[Out]

x^2*(e/(2*b) - (a*f)/(2*b^2)) - c/(a*x) + (f*x^5)/(5*b) + (log(b^(1/3)*x + a^(1/3))*(b^3*c - a^3*f - a*b^2*d +
a^2*b*e))/(3*a^(4/3)*b^(8/3)) - (log(3^(1/2)*a^(1/3)*1i + 2*b^(1/3)*x - a^(1/3))*((3^(1/2)*1i)/2 + 1/2)*(b^3*
c - a^3*f - a*b^2*d + a^2*b*e))/(3*a^(4/3)*b^(8/3)) + (log(3^(1/2)*a^(1/3)*1i - 2*b^(1/3)*x + a^(1/3))*((3^(1/
2)*1i)/2 - 1/2)*(b^3*c - a^3*f - a*b^2*d + a^2*b*e))/(3*a^(4/3)*b^(8/3))

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sympy [A]  time = 4.72, size = 408, normalized size = 1.80 \[ x^{2} \left (- \frac {a f}{2 b^{2}} + \frac {e}{2 b}\right ) + \operatorname {RootSum} {\left (27 t^{3} a^{4} b^{8} + a^{9} f^{3} - 3 a^{8} b e f^{2} + 3 a^{7} b^{2} d f^{2} + 3 a^{7} b^{2} e^{2} f - 3 a^{6} b^{3} c f^{2} - 6 a^{6} b^{3} d e f - a^{6} b^{3} e^{3} + 6 a^{5} b^{4} c e f + 3 a^{5} b^{4} d^{2} f + 3 a^{5} b^{4} d e^{2} - 6 a^{4} b^{5} c d f - 3 a^{4} b^{5} c e^{2} - 3 a^{4} b^{5} d^{2} e + 3 a^{3} b^{6} c^{2} f + 6 a^{3} b^{6} c d e + a^{3} b^{6} d^{3} - 3 a^{2} b^{7} c^{2} e - 3 a^{2} b^{7} c d^{2} + 3 a b^{8} c^{2} d - b^{9} c^{3}, \left (t \mapsto t \log {\left (\frac {9 t^{2} a^{3} b^{5}}{a^{6} f^{2} - 2 a^{5} b e f + 2 a^{4} b^{2} d f + a^{4} b^{2} e^{2} - 2 a^{3} b^{3} c f - 2 a^{3} b^{3} d e + 2 a^{2} b^{4} c e + a^{2} b^{4} d^{2} - 2 a b^{5} c d + b^{6} c^{2}} + x \right )} \right )\right )} + \frac {f x^{5}}{5 b} - \frac {c}{a x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x**9+e*x**6+d*x**3+c)/x**2/(b*x**3+a),x)

[Out]

x**2*(-a*f/(2*b**2) + e/(2*b)) + RootSum(27*_t**3*a**4*b**8 + a**9*f**3 - 3*a**8*b*e*f**2 + 3*a**7*b**2*d*f**2
+ 3*a**7*b**2*e**2*f - 3*a**6*b**3*c*f**2 - 6*a**6*b**3*d*e*f - a**6*b**3*e**3 + 6*a**5*b**4*c*e*f + 3*a**5*b
**4*d**2*f + 3*a**5*b**4*d*e**2 - 6*a**4*b**5*c*d*f - 3*a**4*b**5*c*e**2 - 3*a**4*b**5*d**2*e + 3*a**3*b**6*c*
*2*f + 6*a**3*b**6*c*d*e + a**3*b**6*d**3 - 3*a**2*b**7*c**2*e - 3*a**2*b**7*c*d**2 + 3*a*b**8*c**2*d - b**9*c
**3, Lambda(_t, _t*log(9*_t**2*a**3*b**5/(a**6*f**2 - 2*a**5*b*e*f + 2*a**4*b**2*d*f + a**4*b**2*e**2 - 2*a**3
*b**3*c*f - 2*a**3*b**3*d*e + 2*a**2*b**4*c*e + a**2*b**4*d**2 - 2*a*b**5*c*d + b**6*c**2) + x))) + f*x**5/(5*
b) - c/(a*x)

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